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And all the other braking forces required to turn kinetic energy into heat get magically turned into pixie dust? The mechanical advantage of the braking system varies between disc and rim brakes. The kinetic energy which needs to be transferred into heat does not. Unless your rims are connected to your hubs via magic pixie dust or unicorn hair, all that energy transfer has to take place via the spokes.
Brommers-
Energy is not the question in this case but yeah, I think you're right. The bike slows down through backwards force from the tyre on the ground. That force must pass through to the axle to slow the frame/rider and therefore must always be transmitted through the spokes. Torque/force at the tyre at the point of skidding is always the same thus stress on spokes is always the same.
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The kinetic energy removed maybe the same but due to the rotor size difference between a rim and the disc brake rotor the forces on the spokes are quite different. The higher forces on the spokes applied at the hub flange (set by the hub flange diameter to rim diameter) do not mean higher retardation force on the bike. I have corrected my first post on this. There is a glaring error.
As I have shown torque = force applied * diameter of rotor.
Energy in one revolution is actually work done=torque*2*pi
Assuming tangential spokes
So W.D = (braking torque)2pi = force(rotor diameter). That could be rim or brake.The work done here is fixed. The rotor diameter is not. It's 10 times larger for the rim than the hub flange pcd ( assuming 62mm Shimano 6 bolt flanges) so the force on the spokes has to becl different. Small disc brake hubs just make matters worse.
So bringing energy into this changes nothing.
So when you apply physics/naths the answer falls out. I used to teach this subject and maths too.
The problem is you are misunderstanding the relationship between torque and energy.
frankenbike
cycleclinic
Disc brakes do place more braking loads on spokes than rim brakes can. You can work out what the difference is by taking the ratio of the rim diameter to the rotor diameter.
Force(rim) x d(rim) =force(rotor) *d(rotor)
So for a 160mm rotor a disc brake thd braking torque goes through the hub fglsnge. Braking forces on spokes are therefore 10 times as much load on the spokes as a rim brake can at the limit of tyre adhesion.